Electronic Journal of Qualitative Theory of Differential Equations 2005, No. 21, 1-16;http://www.math.u-szeged.hu/ejqtde/
FIRST ORDER INTEGRO-DIFFERENTIAL EQUATIONS IN BANACH ALGEBRAS INVOLVING CARATHEODORY
AND DISCONTINUOUS NONLINEARITIES
B. C. Dhageand B. D. Karande
Kasubai, Gurukul Colony, Ahmedpur-413 515, Dist: Latur Maharashtra, India e-mail: [email protected]
Abstract
In this paper some existence theorems for the first order differential equations in Banach algebras is proved under the mixed generalized Lipschitz, Carath´eodory and monotonic-ity conditions.
Key words and phrases : Banach algebra, second order differential equation, existence theorem.
AMS (MOS) Subject Classifications: 34K10.
1
Introduction
Let R denote the real line. Given a closed and bounded interval J = [0, T] in R, consider the first order functional integro-differential equation ( in short IGDE)
d dt
h x(t) f(t, x(t))
i =
Z t
0
g(s, x(s))ds, a.e. t∈J
x(0) =x0 ∈R,
(1.1)
wheref :J×R→R− {0}, andg:J×R→R.
By a solution of IGDE (1.1) we mean a functionx∈AC1(J,R) such that
(i) the functiont→ x f(t, x)
is continuous for each x∈R, and
(ii) x satisfies the equations in (1.1),
where, AC1(J,R) is the space of absolutely continuous real-valued functions on J.
conditions. We will employ the fixed point theorems of Dhage [2, 3, 4] for proving our main existence results. The nonlinear differential equation as well as the existence results of this are new to the literature on the theory of ordinary differential equations.
Our method of study is to convert the IGDE (1.1) into equivalent integral equation and apply the fixed point theorems of Dhage [2, 3, 4] under suitable conditions on the nonlinearities f and g. In the following section we shall give some preliminaries needed in the sequel.
2
Auxiliary Results
LetX be a Banach algebra with normk · k. A mappingA:X→X is called D-Lipschitz if there exists a continuous nondecreasing function ψ:R+→R+ satisfying
kAx−Ayk ≤ψ(kx−yk) (2.1)
for all x, y∈ X with ψ(0) = 0. In the special case when ψ(r) =αr (α >0), A is called a Lipchitz with a Lipschitz constantα. In particular, ifα <1,A is called a contraction with a contraction constant α. Further, if ψ(r) < r for all r > 0, then A is called a nonlinear contraction on X. Sometimes we call the function ψ a D-function for convenience.
An operator T : X → X is called compact if T(S) is a compact subset of X for any S ⊂ X. Similarly T : X → X is called totally bounded if T maps a bounded subset of X into the relatively compact subset of X. Finally T : X → X is called completely continuous operator if it is continuous and totally bounded operator on X. It is clear that every compact operator is totally bounded, but the converse may not be true. The nonlinear alternative of Schaefer type recently proved by Dhage [4] is embodied in the following theorem.
Theorem 2.1 (Dhage[4]) Let X be a Banach algebra and letA, B :X→X be two opera-tors satisfying
(a) A is a D-Lipschitz with a D-function ψ,
(b) B is compact and continuous, and
(c) M ψ(r)< r whenever r >0, where M =kB(X)k= sup{kBxk:x∈X}. Then either
(i) the equationλAx Bx=x has a solution for λ= 1, or
(ii) the set E={u∈X|λAu Bu=u,0< λ <1} is unbounded.
It is known that Theorem 2.1 is useful for proving the existence theorems for the integral equations of mixed type. See [2] and the references therein. The method is commonly known as priori bound methodfor the nonlinear equations. See, for example, Dugundji and Granas [7], Zeidler [12] and the references therein.
Corollary 2.1 LetXbe a Banach algebra and letA, B:X →Xbe two operators satisfying (a) A is Lipschitz with a Lipschitz constant α,
(b) B is compact and continuous, and
(c) αM <1 , where M =kB(X)k:= sup{kBxk:x∈X}.
Then either
(i) the equationλAx Bx=x has a solution for λ= 1, or
(ii) the set E={u∈X|λAu Bu=u,0< λ <1} is unbounded.
A non-empty closed set K in a Banach algebra X is called a coneif (i) K+K ⊆K, (ii) λK ⊆ K for λ ∈ R, λ ≥ 0 and (iii) {−K} ∩K = 0, where 0 is the zero element of X. A cone K is called to be positive if (iv) K ◦K ⊆ K, where ”◦” is a multiplication composition in X. We introduce an order relation ≤ inX as follows. Let x, y ∈X. Then x≤yif and only ify−x∈K.A coneKis called to benormalif the normk·kis monotone increasing on K.It is known that if the coneK is normal inX, then every order-bounded set in X is norm-bounded. The details of cones and their properties appear in Guo and Lakshmikantham [9].
We equip the spaceAC1(J,R) with the order relation≤with the help of the cone defined by
K={x ∈C(J,R) :x(t)≥0,∀t∈J}. (2.2) It is well known that the coneKis positive and normal inAC1(J,R).As a result of positivity of the cone K inAC1(J,R) we have:
Lemma 2.1 (Dhage [3]). Let u1, u2, v1, v2 ∈ K be such that u1 ≤ v1 and u2 ≤v2. Then u1u2 ≤v1v2.
For anya, b∈X=AC1(J,R), a≤b,the order interval [a, b] is a set in X defined by
[a, b] ={x∈X:a≤x≤b}. (2.3)
We use the following fixed point theorem of Dhage [3] for proving the existence of extremal solutions of the IGDE (1.1) under certain monotonicity conditions.
Theorem 2.2 (Dhage [3]). Let K be a cone in a Banach algebra X and let a, b ∈ X.
Suppose that A, B: [a, b]→K are two operators such that
(a) A is Lipschitz with a Lipschitz constant α,
(b) B is completely continuous,
(d) A and B are nondecreasing.
Further if the cone K is positive and normal, then the operator equation Ax Bx=x has a least and a greatest positive solution in [a, b], whenever αM <1, where M =kB([a, b])k:= sup{kBxk:x∈[a, b]}.
Theorem 2.3 (Dhage [4]). Let K be a cone in a Banach algebra X and let a, b ∈ X.
Suppose that A, B: [a, b]→K are two operators such that
(a) A is completely continuous,
(b) B is totally bounded,
(c) Ax By∈[a, b] for each x, y∈[a, b],and
(d) B is nondecreasing.
Further if the cone K is positive and normal, then the operator equation Ax Bx=x has a least and a greatest positive solution in [a, b].
Theorem 2.4 (Dhage [4]). Let K be a cone in a Banach algebra X and let a, b ∈ X.
Suppose that A, B: [a, b]→K are two operators such that
(a) A is Lipschitz with a Lipschitz constant α,
(b) B is totally bounded,
(c) Ax By∈[a, b] for each x, y∈[a, b],and
(d) B is nondecreasing.
Further if the cone K is positive and normal, then the operator equation Ax Bx = x has least and a greatest positive solution in [a, b], whenever αM <1, where M =kB([a, b])k:= sup{kBxk:x∈[a, b]}.
Remark 2.1 Note that hypothesis (c) of Theorems 2.2, 2.3, and 2.4 holds if the operatorsA
andB are monotone increasing and there exist elementsaand bin X such thata≤Aa Ba
and Ab Bb≤b.
3
Existence Theory
Let M(J,R) and B(J,R) respectively denote the spaces of measurable and bounded real-valued functions on J. Let C(J,R),be the space of all continuous real-valued functions on J. Define a normk · kinC(J,R) by
kxk= sup
t∈J
ClearlyC(J,R) becomes a Banach algebra with this norm and the multiplication “·” defined by (xy)(t) = x(t)y(t) for all t ∈J. By L1(J,R) we denote the set of Lebesgue integrable functions onJ and the norm k · kin L1(J,R) is defined by
kxkL1 =
Z 1
0
|x(t)|ds.
We need the following lemma in the sequel.
Lemma 3.1 If h∈L1(J,R), then x is a solution of the IGDE
d dt
h x(t) f(t, x(t))
i =
Z t
0
h(s)ds, a.e. t∈J
x(0) =x0,
(3.1)
if and only if is solution of the integral equation (in short IE)
x(t) =
f(t, x(t)) x0 f(0, x0)
+ Z t
0
(t−s)h(s)ds, t∈J. (3.2)
Proof. The proof is simple and omit the details. We need the following definition in the sequel.
Definition 3.1 A mapping β :J×R→Ris said to be Carath´eodory if (i) t→β(t, x) is measurable for eachx∈R, and
(ii) x→β(t, x) is continuous almost everywhere fort∈J.
Again a Carath´odory function β(t, x) is calledL1-Carath´eodory if
(iii) for each real numberr >0 there exists a function hr∈L1(J,R) such that
|β(t, x)| ≤hr(t), a.e. t∈J
for all x∈R with|x| ≤r.
Finally a Carath´eodory function β(t, x) is calledL1X-Carath´eodory if
(iv) there exists a function h∈L1(J,R) such that
|β(t, x)| ≤h(t), a.e. t∈J
for all x∈R.
For convenience, the function h is referred to as a bound function of β.
(H1) The function f : J ×R → R is continuous and there exists a function k ∈ B(J,R)
such thatk(t)>0,a.e. t∈J and
|f(t, x)−f(t, y)| ≤k(t)|x−y|, a.e. t∈J
for all x, y∈R.
(H2) The function gis L1X-Carath´eodory with bound function h.
(H3) There exists a continuous and nondecreasing function Ω : [0,∞) → (0,∞) and a
functionγ ∈L1(J,R) such thatγ(t)>0,a.e. t∈J and
|g(t, x)| ≤γ(t)Ω |x|
, a.e. t∈J,
for all x∈R.
Theorem 3.1 Assume that the hypotheses (H1)-(H3) hold. Suppose that
Z ∞
C1
ds
Ω(s) > C2kγkL1, (3.3)
where
C1 =
F
x0 f(0,x0)
1− kkk
x
0 f(0,x0)
+TkhkL1
, C2=
F T 1− kkk[
x
0 f(0,x0)
+TkhkL1]
,
kkk x
0 f(0,x0)
+TkhkL1
< 1, F = max
t∈J |f(t,0)|, and kkk = maxt∈J |k(t)|. Then the IGDE (1.1) has a solution on J.
Proof. By Lemma 3.1, the IGDE (1.1) is equivalent to integral equation
x(t) =
f(t, x(t)) x0 f(0, x0)
+ Z t
0
(t−s)g(s, x(s))ds, t∈J. (3.4)
SetX =C(J,R). Define the two mappingsA and B onX by
Ax(t) =f(t, x(t)), t∈J, (3.5)
and
Bx(t) = x0 f(0, x0)
+ Z t
0
(t−s)g(s, x(s))ds, t∈J. (3.6)
Obviously A and B define the operators A, B : X → X. Then the IGDE (1.1) is equivalent to the operator equation
We shall show that the operators A andB satisfy all the hypotheses of Corollary 2.1. We first show thatA is a Lipschitz on X.Let x, y∈X. Then by (H1),
|Ax(t)−Ay(t)| ≤ |f(t, x(t))−f(t, y(t))| ≤ k(t)|x(t)−y(t)| ≤ kkk kx−yk
for all t∈J. Taking the supremum over t we obtain
kAx−Ayk ≤ kkkkx−yk
for allx, y∈X.So Ais a Lipschitz onX with a Lipschitz constantkkk.Next we show that B is completely continuous onX. Using the standard arguments as in Granas et al. [8], it is shown that B is a continuous operator on X. Let S be a bounded set in X. We shall show that B(X) is a uniformly bounded and equicontinuous set in X. Since g(t, x(t)) is L1X-Carath´eodory, we have
wherep(t) =T Z t
0
h(s)ds.Therefore,
|Bx(t)−Bx(τ)| →0 as t→τ.
where,
Then u(t)≤w(t) and a direct differentiation ofw(t) yields
w′(t) ≤ C
A change of variables in the above integral gives that
Z w(t)
Now an application of mean value theorem yields that there is a constant M >0 such that w(t)≤M for allt∈J.This further implies that
|x(t)| ≤u(t)≤w(t)≤M.
for allt∈J. Thus the conclusion (ii) of Corollary 2.1 does not hold. Therefore the operator equation AxBx=x and consequently the IGDE (1.1) has a solution onJ. This completes the proof.
4
Existence of Extremal Solutions
We need the following definitions in the sequel.
Definition 4.1 A function u∈AC1(J,R) is called a lower solution of the IGDE (1.1) on
4.1 Carath´eodory case
We consider the following set of assumptions:
(B0) f :J ×R→R+− {0} ,g:J×R→R+ and
x0 f(0, x0)
≥0.
(B1) g is Carath´eodory.
(B2) The functions f(t, x) and g(t, x) are nondecreasing in x andy almost everywhere for
t∈J.
(B3) The IGDE (1.1) has a lower solution u and an upper solutionv onJ with u≤v.
(B4) The function ℓ:J →Rdefined by
ℓ(t) =|g(t, u(t))|+|g(t, v(t))|, t∈J
is Lebesgue measurable.
Remark 4.1 Assume that (B2)-(B4) hold. Then
|g(t, x(t))| ≤ℓ(t), a.e. t∈J,
for all x∈[u, v].
Theorem 4.1 Suppose that the assumptions (H1)-(H3) and (B0)-(B4 ) hold. Further if
kkk f x0
(0,x0)
+TkℓkL1
<1, andℓ is given in Remark 4.1, then IGDE (1.1) has a minimal and a maximal positive solution on J.
Proof. Now IGDE (1.1) is equivalent to IE (3.4) onJ.LetX=C(J,R) and define an order relation “≤” by the coneK given by (2.2). ClearlyK is a normal cone inX. Define two operatorsA andB onX by (3.5) and (3.5) respectively. Then IE (1.1) is transformed into an operator equation Ax(t)Bx(t) = x(t) in a Banach algebra X. Notice that (B1) implies
A, B: [u, v]→K.Since the coneK inXis normal, [u, v] is a norm bounded set in X.Now it is shown, as in the proof of Theorem 3.1, thatA is a Lipschitz with a Lipschitz constant kαk and B is completely continuous operator on [u, v].Again the hypothesis (B2) implies
that A and B are nondecreasing on [u, v]. To see this, let x, y∈ [u, v] be such that x ≤y. Then by (B2),
Ax(t) =f(t, x(t))≤f(t, y(t)) =Ay(t), ∀t∈J. Similarly,
Bx(t) = x0 f(0, x0)
+ Z t
0
(t−s)g(s, x(s))ds
≤ x0 f(0, x0)
+ Z t
0
(t−s)g(s, x(s))ds
for all t ∈ J. So A and B are nondecreasing operators on [u, v]. Again Lemma 4.1 and hypothesis (B3) implies that
u(t) ≤ [f(t, u(t))]
x0
f(0, x0)
+ Z t
0
(t−s)g(s, u(s)))ds
≤ [f(t, x(t))]
x0
f(0, x0)
+ Z t
0
(t−s)g(s, x(s))ds
≤ [f(t, v(t))]
x0
f(0, x0)
+ Z t
0
(t−s)g(s, v(s)))ds
≤ v(t),
for all t ∈ J and x ∈ [u, v]. As a result u(t) ≤ Ax(t)Bx(t) ≤ v(t),∀t ∈ J and x ∈ [u, v]. Hence Ax Bx∈[u, v] for all x∈[u, v].
Again
M = kB([u, v])k
= sup{kBxk:x∈[u, v]}
≤ sup
x0
f(0, x0)
+Tsup
t∈J
Z t
0
|g(s, x(s))|ds x∈[u, v]
≤
x0
f(0, x0)
+T
Z T
0
ℓ(s)ds
=
x0 f(0, x0)
+TkℓkL1.
SinceαM ≤ kkk( f x0
(0,x0)
+TkℓkL1)<1,we apply Theorem 4.1 to the operator equation
Ax Bx=xto yield that the IGDE (1.1) has a minimal and a maximal positive solution on J. This completes the proof.
4.2 Discontinuous case
We need the following definition in the sequel.
Definition 4.3 A mapping β :J×R→Ris said to be Chandrabhan if (i) t→β(t, x) is measurable for eachx∈R, and
(ii) x→β(t, x) is nondecreasing almost everywhere for t∈J.
Again a Chandrabhan function β(t, x) is calledL1-Chandrabhan if
(iii) for each real numberr >0 there exists a function hr∈L1(J,R) such that
|β(t, x)| ≤hr(t), a.e. t∈J
Finally a Chandrabhan function β(t, x) is called L1X-Chandrabhan if
(iv) there exists a function h∈L1(J,R) such that
|β(t, x)| ≤h(t), a.e. t∈I
for all x∈R.
For convenience, the function h is referred to as a bound function of β.
We consider the following hypotheses in the sequel.
(C1) The function f is continuous on J×R.
(C2) There is a function k∈B(J,R) such that k(t)>0,a.e. t∈I and
|f(t, x)−f(t, y)| ≤k(t)|x−y|, a.e. t∈J
for all x, y∈R.
(C3) The function f(t, x) is nondecreasing in x almost everywhere fort∈J.
(C4) The function gis Chandrabhan.
Theorem 4.2 Suppose that the assumptions (B0),(B3)-(B4) and (C1)-(C4) hold. Then IGDE (1.1) has a minimal and a maximal positive solution on J.
Proof. Now IGDE (1.1) is equivalent to IE (3.4) on J. Let X = C(J,R) and define an order relation “≤” by the coneKgiven by (2.2). ClearlyKis a normal cone inX. Define two operatorsAandB onXby (3.5) and (3.6) respectively. Then FIE (1.1) is transformed into an operator equation Ax(t)Bx(t) = x(t) in a Banach algebra X. Notice that (B0)
implies A, B: [u, v]→K.Since the coneK inX is normal, [u, v] is a norm bounded set in X.
Step I : Next we show thatAis completely continuous on [a, b]. Now the coneK inX is normal, so the order interval [a, b] is norm-bounded. Hence there exists a constantr >0 such thatkxk ≤r for all x∈[a, b]. Asf is continuous on compactJ×[−r, r], it attains its maximum, say M. Therefore for any subset S of [a, b] we have
kA(S)kP = sup{kAxk:x∈S}
= supnsup
t∈J
|f(t, x(t))|:x∈So
≤supnsup
t∈J
|f(t, x)|:x∈[−r, r]o
≤M.
Next we note that the functionf(t, x) is uniformly continuous on [0, T]×[−r, r]. There-fore for anyt, τ ∈[0, T] we have
|f(t, x)−f(τ, x)| →0 as t→τ
for all x∈[−r, r]. Similarly for any x, y∈[−r, r]
|f(t, x)−f(t, y)| →0 as x→y
for all t∈[0, T]. Hence any t, τ ∈[0, T] and for any x∈S one has
|Ax(t)−Ax(τ)|=|f(t, x(t))−f(τ, x(τ))|
≤ |f(t, x(t))−f(τ, x(t)|+|f(τ, x(t))−f(τ, x(τ))| →0 as t→τ.
This shows that A(S) is an equi-continuous set inX. Now an application of Arzel`a-Ascoli theorem yields that A is a completely continuous operator on [a, b].
Step II : Next we show that B is totally bounded operator on [a, b]. To finish, we shall show that B(S) is uniformly bounded ad equi-continuous set in X for any subset S of [a, b]. Since the cone K inX is normal, the order interval [a, b] is norm-bounded. Hence there is a real numberr >0 such thatkxk ≤r for allx∈[a, b]. Lety∈B(S) be arbitrary. Then,
y(t) = x0 f(0, x0)
+ Z t
0
(t−s)g(s, x(s))ds
for some x∈S. By hypothesis (B2) one has
|y(t)|=
x0
f(0, x0)
+T
Z t
0
|g(s, x(s))|ds
≤
x0 f(0, x0)
+T
Z t
0
ℓ(s)ds
≤
x0 f(0, x0)
+TkℓkL1.
Taking the supremum over t,
kyk ≤
x0
f(0, x0)
which shows that B(S) is a uniformly bounded set inX. Similarly let t, τ ∈J. Then for
ℓ(s)ds. Since the function p is continuous on compact interval J, it is uniformly continuous , and therefore we have
|y(t)−y(τ)| →0 as t→τ,
for ally∈B(S). Hence B(S) is an equi-continuous set inX. ThusB is totally bounded in view of Arzel`a-Ascoli theorem.
Thus all the conditions of Theorem 2.3 are satisfied and hence an application of it yields that the IGDE (1.1) has a maximal and a minimal solution onJ.
Theorem 4.3 Suppose that the assumptions (B0),(B3) and (C1)-(C4) hold. Further if and a maximal positive solution on J.
Proof. Now IGDE (1.1) is equivalent to IE (3.4) on J. Let X = C(J,R) and define an order relation “≤” by the coneKgiven by (2.2). ClearlyKis a normal cone inX. Define two operatorsAandB onXby (3.5) and (3.6) respectively. Then FIE (1.1) is transformed into an operator equation Ax(t)Bx(t) = x(t) in a Banach algebra X. Notice that (B0)
implies A, B : [u, v] → K. Since the cone K in X is normal, [u, v] is a norm bounded set in X. Now it can be shown as in the proofs of Theorem 3.1 and Theorem 2.4 that the operator Ais a Lipschitz with a Lipschitz constant α=kkk andB is totally bounded with M =kB([u, v])k = the desired conclusion follows by an application of Theorem 2.4. This completes the proof.
5
An Example
Given the closed and bounded intervalJ = [0,1] in R,consider the nonlinear IGDE
d dt
h x(t) f(t, x(t))
i =
Z t
0
p(s)x(s) 1 +|x(s)|
ds, a.e. t∈J (5.1)
wherep∈L1(J,R) and f :J×R→Ris defined by
f(t, x) = 1 2
1 +α|x|
, α >0
for all t∈J. Obviouslyf :J×R→R+− {0}.Define a functiong:J×R→Rby
g(t, x) = p(t)x 1 +|x|.
It is easy to verify that f is continuous and Lipschitz on J ×Rwith a Lipschitz constant α.Furtherg(t, x) isL1X-Carath´eodory with the bound functionh(t) =p(t) onJ. Therefore if α(1 +kpkL1) < 1, then by Theorem 3.1, the IGDE (5.1) has a solution on J, because
the function Ω satisfies condition (3.3) with γ(t) =p(t) for all t ∈J and Ω(r) = 1 for all r ∈R+.
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